Four charges q, 2q, -4q and 2q are placed in | vedclass

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(B) Let the corners of the square be $A, B, C, D$ in order. Let the charges be $q_A = q, q_B = 2q, q_C = -4q, q_D = 2q$.The distance from each corner

Vedclass · Accepted Answer

$\frac{5q}{2\pi \varepsilon_0 b^2}$ from $+q$ to $-4q$

Vedclass · Answer

(B) Let the corners of the square be $A, B, C, D$ in order. Let the charges be $q_A = q, q_B = 2q, q_C = -4q, q_D = 2q$.The distance from each corner to the centre $O$ is $r = \frac{b}{\sqrt{2}}$.The electric field at the centre due to a charge $Q$ is $E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{Q}{b^2/2} = \frac{Q}{2\pi \varepsilon_0 b^2}$.Field due to $q_A$ at $O$: $\vec{E}_A = \frac{q}{2\pi \varepsilon_0 b^2}$ directed away from $A$ (towards $C$).Field due to $q_C$ at $O$: $\vec{E}_C = \frac{4q}{2\pi \varepsilon_0 b^2}$ directed towards $C$.Net field along $AC$: $\vec{E}_{AC} = \vec{E}_A + \vec{E}_C = \frac{q + 4q}{2\pi \varepsilon_0 b^2} = \frac{5q}{2\pi \varepsilon_0 b^2}$ towards $C$.Field due to $q_B$ at $O$: $\vec{E}_B = \frac{2q}{2\pi \varepsilon_0 b^2}$ directed away from $B$ (towards $D$).Field due to $q_D$ at $O$: $\vec{E}_D = \frac{2q}{2\pi \varepsilon_0 b^2}$ directed away from $D$ (towards $B$).Net field along $BD$: $\vec{E}_{BD} = \vec{E}_B - \vec{E}_D = 0$.The resultant field is $\vec{E}_{net} = \vec{E}_{AC} = \frac{5q}{2\pi \varepsilon_0 b^2}$ directed from $A$ to $C$. Since $A$ has charge $+q$ and $C$ has charge $-4q$,the direction is from $+q$ to $-4q$.

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

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